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Did some testing with a CAS and found this cute fact:

Consider three points $a$, $b$, $c$ in the complex plane, and $e$ the center of the nine point circle of the triangle $a b c$. Show that

$$e = \frac{1}{2}\cdot \frac{\left| \begin{matrix} 1& a^2 & \bar a \\ 1& b^2 & \bar b \\ 1& c^2 & \bar c \end{matrix}\right| } {\left | \begin{matrix} 1 & a & \bar a \\ 1& b & \bar b \\ 1& c & \bar c \end{matrix}\right| }$$

This is all probably known, but haven't seen it before. Any feedback, or references would be appreciated.

orangeskid
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    Have you tried proving it algebraically? What have you found? I'd start with $\left|\frac{a+b}{2}-e\right| = \left|\frac{b+c}{2}-e\right| = \left|\frac{c+a}{2}-e\right|$ and see what I can get from that. – aschepler Jan 13 '24 at 19:46

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I don't have a complete exposition, but I think an approach to a proof might be as follows:

Since the nine-point circle passes through the midpoints of each side, it is the circumscribed circle of the triangle that joins these midpoints, which would have representations $\{\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\}$. Then we would presumably use the result derived here to complete the proof.

heropup
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